AP EAMCET202119 Aug 2021Evening ShiftPhysicsRotational MotionActual
A girl of mass M stands on the rim of a friction less merry-go-round, of radius R and rotational inertia I , that is not moving. She throws a rock of mass m horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground is v . Afterwards, the linear speed of the girl is
Options
- Am v R 2 I + M R 2
- B( m + M ) v R 2 I + M R 2
- Cm v R 2 I + ( M + m ) R 2
- Dm v R 2 I + ( M − m ) R 2
Correct answer
A. m v R 2 I + M R 2
Step-by-step solution
The initial angular momentum of the system is zero. The final angular momentum of the girls-plus-merry-go-round is I + M R 2 ω The final angular momentum we associate with the thrown rock is negative ⇒ - m R v ⇒ m R v = I + M R 2 ω ⇒ m R v I + M R 2 = ω ⇒ R ω = m v R 2 I + M R 2